An electric heater rated at 200\text{ W} is used to heat a 0.40\text{ kg} sample of a pure… — JAMB UTME

An electric heater rated at $200\text{ W}$ is used to heat a $0.40\text{ kg}$ sample of a pure solid substance maintained at its melting point. If it takes $2\text{ minutes}$ for exactly half of the sample to melt into liquid at the same temperature, what is the specific latent heat of fusion of the substance?

$1.2 \times 10^5\text{ J kg}^{-1}$ Answer
B
$6.0 \times 10^4\text{ J kg}^{-1}$
C
$2.0 \times 10^3\text{ J kg}^{-1}$
D
$1.2 \times 10^7\text{ J kg}^{-1}$

Answer

The specific latent heat of fusion of the substance is

1.2 \times 10^5\text{ J kg}^{-1}

The thermal energy absorbed during melting is

Q = P \times t = 200\text{ W} \times 120\text{ s} = 24,000\text{ J}

. Since only half of the

0.40\text{ kg}

sample melts, the mass undergoing phase change is

0.20\text{ kg}

. Dividing the heat absorbed (

24,000\text{ J}

) by the melted mass (

0.20\text{ kg}

) yields the correct specific latent heat of fusion of

1.2 \times 10^5\text{ J kg}^{-1}

Step-by-Step Solution

Calculate total thermal energy supplied by the heater.

Q = P \times t = 200\text{ W} \times (2 \times 60\text{ s}) = 24,000\text{ J}

Electrical energy supplied is converted completely into thermal energy.

Determine the mass of substance that melted.

m_{\text{melted}} = \frac{1}{2} \times 0.40\text{ kg} = 0.20\text{ kg}

Only half of the total mass undergoes phase change.

Compute the specific latent heat of fusion

L_f

L_f = \frac{Q}{m_{\text{melted}}} = \frac{24,000\text{ J}}{0.20\text{ kg}} = 120,000\text{ J kg}^{-1} = 1.2 \times 10^5\text{ J kg}^{-1}

Phase change occurs at constant temperature using the latent heat equation

Q = m L_f

Key Concept

Latent Heat of Fusion and Energy Balance

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